Polar Coordinates; Vectors

Transcription

1 Chapter 10 Polar Coordinates; Vectors 10.R Chapter Review 1. 3, 6 x = 3cos 6 = , 3 x = 4cos 3 = y = 3sin 6 = 3 3 3, 3 y =4sin 3 = 3 (, 3) 3., 4 3 x = cos 4 3 =1 4. 1, 5 4 x = 1cos 5 4 = y = sin 4 3 = 3 ( 1, 3 ) y = 1sin 5 4 =, 5. 3, x = 3cos = 0 y = 3sin = 3 (0, 3) 6. 4, 4 x = 4cos 4 = y= 4sin 4 = (, ) 7. The point ( 3, 3) lies in quadrant II. r = x + y = ( 3) + 3 = 3 = tan 1 y = tan 1 3 = tan 1 ( 1) = x 3 4 Polar coordinates of the point ( 3, 3 ) are 3, 4 or 3,

2 Section 10.R Chapter Review 8. The point (1, 1) lies in quadrant IV. r = x + y = 1 + ( 1) = = tan 1 y = tan 1 1 = tan 1 ( 1) = x 1 4 Polar coordinates of the point ( 1, 1 ) are, 4 and, The point (0, ) lies on the negative y-axis. r = x + y = 0 + ( ) = = tan 1 y = tan 1 = x 0 Polar coordinates of the point ( 0, ) are, or,. 10. The point (, 0) lies on the positive x-axis. r = x + y = + 0 = 4 = = tan 1 y = tan 1 0 = tan 1 0 = 0 x Polar coordinates of the point (, 0) are (, 0) and (, ). 11. The point (3, 4) lies in quadrant I. r = x + y = = 5 = tan 1 y = tan x 3 Polar coordinates of the point ( 3, 4) are ( 5, 0.93) or ( 5, 4.07). 1. The point ( 5, 1) lies in quadrant II. r = x + y = ( 5) + 1 =13 = tan 1 y = tan x 5 Polar coordinates of the point ( 5, 1) are ( 13, 1.96) or ( 13, 5.10) x + 3y = 6y x + y = y r = rsin r rsin = x y = 5y r cos r sin = 5rsin ( ) = 5r sin r cos sin r cos ( ) = 5sin 15. x y = y x 16. x + y = y x x + y 3y = y x x + y x = y x ( ) 3y = y x x + y r 3( rsin ) = tan r 3r sin tan = 0 r ( 3sin ) tan = 0 ( ) x = y x x + y r ( rcos ) = tan r r cos tan = 0 ( ) tan = 0 r cos 1013

3 Chapter 10 ( ) = 4 r cos ( r ) = 4 r 3 cos = x x + y Polar Coordinates; Vectors ( ) = 3 rsin ( r cos r sin ) = 3 r 3 sin ( cos sin ) = y x y r 3 sin cos ( ) = r = sin r = rsin x + y = y x + y y = 0 1. r = 5 r = 5 x + y = 5 3. r cos + 3rsin = 6 x + 3y = r = sin 3r = rsin 3(x + y ) = y 3x + 3y y = 0. = 4 tan = tan 4 y x =1 y = x 4. r tan = 1 x + y = 1 ( ) y x x y + y 3 = x y 3 + x y x = 0 5. r = 4cos The graph will be a circle. Check for symmetry: Polar axis: Replace by. The result is r = 4cos( ) = 4cos. The graph is symmetric with respect to the polar axis. The line = : Replace by. r=4cos( )=4(cos( )cos + sin( )sin ) = 4( cos + 0) = 4cos The test fails. The pole: Replace r by r. r = 4cos. The test fails. Due to symmetry to the polar axis, assign values to from 0 to r = 4cos

4 Section 10.R Chapter Review 6. r = 3sin The graph will be a circle. Check for symmetry: Polar axis: Replace by. The result is r = 3sin( ) = 3sin. The test fails. The line = : Replace by. r=3sin( )=3(sin ( )cos cos( )sin ) = 3(0 + sin ) = 3sin The graph is symmetric with respect to the line =. The pole: Replace r by r. r = 3sin. The test fails. Due to symmetry to the line =, assign values to from to r = 3sin

5 Chapter 10 Polar Coordinates; Vectors 7. r = 3 3sin The graph will be a cardioid. Check for symmetry: Polar axis: Replace by. The result is r = 3 3sin( ) = 3 + 3sin. The test fails. The line = : Replace by. r=3 3sin( ) = 3 3(sin( )cos cos( )sin ) = 3 3(0+ sin ) = 3 3sin The graph is symmetric with respect to the line =. The pole: Replace r by r. r = 3 3sin. The test fails. Due to symmetry to the line =, assign values to from to. 3 r = 3 3sin r = + cos The graph will be a limacon without an inner loop. Check for symmetry: Polar axis: Replace by. The result is r = + cos( ) = + cos. The graph is symmetric with respect to the polar axis. The line = : Replace by. r=+cos( ) = + (cos( )cos + sin( )sin ) = + ( cos + 0) = cos The test fails. The pole: Replace r by r. r = + cos. The test fails. Due to symmetry to the polar axis, assign values to from 0 to. 1016

6 Section 10.R Chapter Review r = + cos r = 4 cos The graph will be a limacon without an inner loop. Check for symmetry: Polar axis: Replace by. The result is r = 4 cos( ) = 4 cos. The graph is symmetric with respect to the polar axis. The line = : Replace by. r=4 cos( ) = 4 (cos( )cos + sin( )sin ) = 4 ( cos + 0) = 4 + cos The test fails. The pole: Replace r by r. r = 4 cos. The test fails. Due to symmetry to the polar axis, assign values to from 0 to r = 4 cos

7 Chapter 10 Polar Coordinates; Vectors 30. r = 1 sin The graph will be a limacon with an inner loop. Check for symmetry: Polar axis: Replace by. The result is r = 1 sin( ) = 1 + sin. The test fails. The line = : Replace by. r=1 sin( ) =1 (sin( )cos cos( )sin ) = 1 (0+ sin ) = 1 sin The graph is symmetric with respect to the line =. The pole: Replace r by r. r = 1 sin. The test fails. Due to symmetry to the line =, assign values to from to r = 1 sin

8 Section 10.R Chapter Review 31. r = x + y = ( 1) + ( 1) = tan = y x = 1 1 = 1 = 5º The polar form of z = 1 i is z = r( cos + isin ) = ( cos( 5º ) + isin( 5º )). 3. r = x + y = ( 3) +1 = 4 = tan = y x = 1 3 = 3 3 = 150º The polar form of z = 3 + i is z = r( cos + isin ) = ( cos( 150º ) + isin( 150º )). 33. r = x + y = 4 + ( 3) = 5 = 5 tan = y x = º The polar form of z = 4 3i is z = r( cos + isin ) = 5 cos 33.1º 34. r = x + y = 3 + ( ) = 13 tan = y x = º ( ( ) + isin( 33.1º )). The polar form of z = 3 i is z = r( cos + isin ) = 13( cos( 36.3º ) + isin( 36.3º )). ( ( ) + isin( 150º )) = cos 150º cos 60º ( ( ) + isin( 60º )) = i = 3+i i = i 1019

9 Chapter 10 Polar Coordinates; Vectors cos 3 + isin 3 = i = i cos isin 3 4 = 4 + i = + i ( ( ) + isin( 350º )) i ( ) = i cos 350º ( ( ) + isin( 160º )) 0.5( i) = i cos 160º 41. z w = ( cos( 80º ) + i sin( 80º )) ( cos( 50º ) + isin( 50º )) =1 1cos(80º+50º)+isin(80º ( + 50º) ) = cos 130º z ( cos( 80º ) + i sin( 80º )) = cos(80º 50º)+isin(80º 50º) w cos 50º ( ) + isin( 130º ) ( ( ) + isin( 50º )) = 1 1 ( ) ( ) + isin( 30º ) = cos 30º 4. z w = ( cos( 05º ) + i sin( 05º )) ( cos( 85º ) + isin( 85º )) =1 1cos(05º+85º)+isin(05º ( + 85º) ) = cos 90º z ( cos( 05º ) + i sin( 05º )) = = 1 cos(05º 85º)+isin(05º 85º) w cos 85º 1 ( ( ) + isin( 85º )) ( )+isin10º ( ) = cos10º ( ) + isin( 90º ) ( ) 43. z w = 3 cos isin 9 5 cos 5 + isin 5 = 3 cos isin = 6 ( cos( ) + isin( ) ) = 6( cos( 0) + isin( 0) ) z 3 cos 9 w = 5 + isin 9 5 cos 5 + isin = 3 cos isin = 3 cos isin z w = cos isin cos 3 + isin 3 = 3 cos i sin = 6 ( cos( ) + isin( ) ) = 6 ( cos( 0) + i sin( 0 ) z cos 5 w = 3 + isin cos 3 + isin = 3 cos isin = 3 cos i sin z w = 5( cos( 10º ) + isin( 10º )) cos 355º ( ( ) + i sin( 355º )) 100

10 Section 10.R Chapter Review z w ( ) = 5 ( cos( 365º ) + isin( 365º )) ( ( ) + isin( 5º )) ( ( ) + i sin( 10º )) ( cos( 355º ) + isin( 355º )) = 5 ( cos(10º 355º) + i sin(10º 355º) ) 1 ( )=5 ( cos15º ( )+isin( 15º )) = 5 1cos(10º+ 355º) + isin(10º+ 355º) = 5 cos 5º 5 cos 10º = = 5 cos( 345º)+ isin( 345º) ( ( ) + isin( 50º )) ( cos( 340º ) + isin( 340º )) ( ) = 4 ( cos( 390º ) + isin( 390º )) ( ( ) + isin( 30º )) ( ( ) + i sin50º ) ( cos( 340º ) + isin( 340º )) ( cos(50º 340º) + i sin(50º 340º) ) 1 ( )=4 ( cos( 70º ) + i sin( 70º )) 46. z w = 4 cos 50º = 4 1cos(50º+340º) + isin(50º + 340º) z w = 4 cos 30º 4 cos 50º = cos 0º = 4 cos( 90º)+ isin( 90º) [ ( ( ) + isin( 0º ))] 3 = 3 3 cos(3 0º) + isin(3 0º) 48. cos 50º ( )=7 cos 60º = i = i [ ( ( ) + isin( 50º ))] 3 = 3 cos(3 50º) + isin(3 50º) 49. cos isin = = 4 cos 5 + i sin 5 ( )=8 cos 150º = i = i ( ) 4 cos isin = 4( i) = 4i ( ( ) + i sin( 60º )) ( ( ) + isin( 150º )) 50. cos 5 + isin 5 4 = 4 cos isin =16 cos isin 5 4 =16 i = 8 8 i i r = i = cos 300º ( 1 3 i) 6 = cos 300º ( ) = tan = 3 ( ( ) + isin( 300º )) 1 = 3 = 300º [ ( ( ) + i sin( 300º ))] 6 = 6 ( cos( 6 300º ) + isin( 6 300º )) ( ( ) + i sin1800º ( ))=64 ( cos( 0º ) + isin( 0º )) = 64 cos 1800º = i =

11 Chapter 10 Polar Coordinates; Vectors 5. i r = + ( ) = tan = i = ( cos( 315º ) + isin( 315º )) ( i) 8 = cos 315º = 1 = 315º [ ( ( ) + isin( 315º ))] = ( ) ( cos( 8 315º ) + i sin( 8 315º )) ( ( ) + isin( 50º )) = 4096 ( cos( 0º ) + i sin( 0º )) = 4096 cos 50º = 4096(1 + 0 i) = i r = = 5 tan = i = 5 cos 53.1º (3 + 4i) 4 = 5 4 cos º 53.1º ( ( ) + isin( 53.1º ) [ ( ( ) + i sin( º ))] = 65 ( cos( 1.4º ) + isin( 1.4º )) ( ) = i i( ) i r = 1 + ( ) = 5 tan = = 96.6º 1 1 i = 5 cos 96.6º (1 i) 4 = 5 ( ( ) + isin( 96.6º )) ( ) 4 cos º = 5 cos º ( ( ) + i sin( º )) ( ( ) + isin( º )) ( ) = i i(0.9593) i r = = 7 tan = 0 7 = 0 = 0º ( ( ) + isin( 0º )) 7 + 0i = 7 cos 0º The three complex cube roots of 7 = 7 cos0º+isin0º 3 z k = 7 cos 0º 360º k + + i sin 0º 360º k = 3 cos 10º k ( ) are: [ ( ) + isin( 10º k) ] [ ( ) + i sin( 10º 0) ] = 3 ( cos( 0º ) + i sin( 0º )) = 3 z 0 = 3 cos 10º 0 [ ( ) + isin( 10º 1) ] = 3 ( cos( 10º ) + isin( 10º )) = z 1 = 3 cos 10º 1 z =3cos 10º 4 i [ ( ) + isin( 10º ) ] = 3 ( cos( 40º ) + isin( 40º )) = i r = 16 ( ) + 0 = 56 = 16 tan = 0 ( ( ) + isin( 180º )) ( ( ) + isin( 180º )) are: 16= 16 cos 180º The four complex fourth roots of 16= 16 cos 180º 16 = 0 = 180º 10

12 Section 10.R Chapter Review 4 z k = 16 cos 180º 4 = cos 45º + 90º k + 360º k 4 + isin 180º º k 4 [ ( ) + i sin( 45º + 90º k) ] [ ( ) + isin( 45º+ 90º 0) ] = ( cos( 45º ) + isin( 45º )) = + i [ ( ) + isin( 45º+ 90º 1) ] = ( cos135º ( )+isin( 135º )) = + i [ ( ) + isin( 45º+ 90º ) ] = ( cos( 5º ) + i sin( 5º )) = i [ ( ) + isin( 45º+ 90º 3) ] = ( cos( 315º ) + isin( 315º )) = i z 0 = cos 45º + 90º 0 z 1 = cos 45º + 90º 1 z =cos 45º + 90º z 3 =cos 45º + 90º P = (1, ), Q = (3, 6) v = (3 1)i + ( 6 ( ))j = i 4j v = +( 4) = 0 = P = ( 3, 1), Q = (4, ) v = (4 ( 3))i + ( 1)j = 7i 3j v = 7 + ( 3) = P = (0, ), Q = ( 1, 1) v = ( 1 0)i + (1 ( ))j = 1i+3j v = ( 1) + 3 = P = (3, 4), Q = (, 0) v = ( 3)i + (0 ( 4))j = 5i+ 4j v = ( 5) + 4 = v = i + j, w = 4i 3j 4v 3w = 4( i + j) 3( 4i 3j) = 8i + 4j 1i + 9j = 0i +13j 6. v = i + j, w = 4i 3j v + w = ( i + j) + ( 4i 3j) = i j + 8i 6j = 10i 7j 63. v = i + j v = i + j = ( ) +1 = v = i + j, w = 4i 3j v + w = i + j 65. v = i + j, w = 4i 3j ( ) + ( 4i 3j) = i j = + ( ) = 8 = v + w = i + j + 4i 3j = ( ) ( 3) = v = i + j, w = 4i 3j v 3 w = ( i + j) 3 4i 3j = 4i + j 3 4i 3j = ( 4) ( 3) = =

13 Chapter 10 Polar Coordinates; Vectors 67. u = v v = i + j i + j = i + j ( ) +1 = i+j 5 = 5 5 i+ 5 5 j 68. u = w w = (4i 3j) 4i 3j = 4i + 3j 4i + 3j = 4 + ( 3) 5 = 4 5 i j 69. v = i + j, w = 4i 3j v w = (4)+1( 3) = 8 3 = 11 cos = v w v w = 11 ( ) ( 3) = = º 70. v = 3i j, w = i + j v w = 3 1+ ( 1)(1) = 3 1 = cos = v w v w = 3 + ( 1) = 10 = º 71. v = i 3j, w = i + j v w = 1( 1) + ( 3)(1) = 1 3 = 4 cos = v w v w = ( 3) ( 1) +1 = 4 10 = º 7. v = i + 4 j, w = 3i j v w = 1(3)+ (4)( )=3 8= 5 cos = v w v w = ( ) = = º 73. proj w v = v w i +3j w = 3i + j w 3 +1 ( ) ( 3i + j) ( ) = ( 3i + j) = i j 74. proj w v = v w ( i + j) ( 3i j) w = 3i j w 3 +( 1) = i j = 3 i + 1 j ( ) = ( 1) 10 ( 3i j) 75. Let the positive x-axis point downstream, so that the velocity of the current is v c = i. Let v w = the velocity of the swimmer in the water. Let v g = the velocity of the swimmer relative to the land. Then v g = v w + v c The speed of the swimmer is v w = 5; its direction is directly across the river, so 104

14 Section 10.R Chapter Review Let v w = 5 j. v g = v w + v c = 5j + i = i + 5j Let v g = + 5 = miles per hour. Since the river is 1 mile wide, it takes the swimmer 0. hours to cross the river. The swimmer will end up (0.)() = 0.4 miles downstream. 76. Let v a = the velocity of the plane in still air. v w = the velocity of the wind. v g = the velocity of the plane relative to the ground. v g = v a + v w v a = 500 j v w = 60 i j = 30 i 30 j v g = v a + v w = 500 j + 30 i 30 j = ( 30 )i The speed of the plane relative to the ground is: v g = 30 ( ) j ( ) + ( ) = = 11, miles per hour To find the direction, find the angle between v g and a convenient vector such as due north, j. v cos = g j ( = 30 ) v g j ( )(1) = º The plane is traveling with a ground speed of about miles per hour in a direction of 5.3 east of north. 77. Let F 1 be the tension on the left cable and F be the tension on the right cable. Let F 3 represent the force of the weight of the box. F 1 = F 1 ( cos( 140º )i + sin( 140º )j) F 1 ( i j) F = F cos( 30º )i + sin( 30º )j ( ) ( ) F i j F 3 = 000j For equilibrium, the sum of the force vectors must be zero. F 1 + F + F 3 = F 1 i F 1 j F i F j 000j = ( F F )i + ( F F 000)j = 0 Set the i and j components equal to zero and solve: 105

15 Chapter 10 Polar Coordinates; Vectors F F = 0 F = F 1 = F F F 000 = F F F 1 = 000 F pounds ( ) 000 = 0 F = (1843) = 1630 pounds The tension in the left cable is about 1843 pounds and the tension in the right cable is about 1630 pounds

16 Section 10.R Chapter Review 79. Let the positive x-axis point downstream, so that the velocity of the current is v c = 5i. Let v w = the velocity of the boat in the water. Let v L = the velocity of the boat relative to the land. Then v L = v w + v c The speed of the boat is v w =15; we need to find the direction. Let v w = ai + b j so v w = a + b =15 a + b = 5. Let v L = k j. Since v L = v w + v c, k j = ai + b j+ 5i k j = (a + 5)i + b j a + 5 = 0 and k = b a = 5 a +b =5 5 + b = 5 b = 00 k = b = 00 v w = 5i+ 00j and v L = 00j Find the angle between v w and v L : ( ) ( ) cos = v v w L = v w v L 15 ( ) 00 = = = cos º The heading of the boat needs to be upstream, in other words, the boat should head at an angle of = o to the shore. v c v w v L Current flowing shore θ α 107